Question

A. Find a vector \( \vec{y} \) that is perpendicular to \( \vec{x} = \begin{bmatrix} 3 \ -1 \end{bmatrix} \). B. Find the angle between \( \vec{x} = \begin{bmatrix} 3 \ -1 \end{bmatrix} \) and \( \vec{y} = \begin{bmatrix} 6 \ 2 \end{bmatrix} \).

          A. Find a vector \( \vec{y} \) that is perpendicular to \( \vec{x} = \begin{bmatrix} 3 \ -1 \end{bmatrix} \). 
B. Find the angle between \( \vec{x} = \begin{bmatrix} 3 \ -1 \end{bmatrix} \) and \( \vec{y} = \begin{bmatrix} 6 \ 2 \end{bmatrix} \).

A. Find a vector y⃗ that is perpendicular to x⃗ =
< b m a t r i x >.
B. Find the angle between x⃗ =
< b m a t r i x > and y⃗ =
< b m a t r i x >.

Added by Amanda C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Aparna Shakti

Step 1

To find a vector that is perpendicular to [3], we can take the cross product of [3] with any other vector. Let's choose the vector [1, 0, 0] as the other vector. Taking the cross product of [3] and [1, 0, 0] gives us: [3] x [1, 0, 0] = [0, 3, 0] So, the vector y Show more…

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